Calculation of the adjusted mean

[thereddevils]

Homework Statement

Four runners A,B,C, and D ran 100m each. The time taken, in seconds by each runner can be considered as independent observation from a normal distribution with mean 14 and SD 0.2 . A runner E ran 400 m. THe time taken by E can be considered as an observation from a normal distribution with mean 58 and SD 1.0 independent of the times taken by the other runners. For the four runners A,B,C and D , find the probabiltiy that A is the fastest runner. Find also the probability that runners C and D (in any order of arrangement) are the slowest runner.

Homework Equations

The Attempt at a Solution

Maybe some hints to get me started on both parts of the questions.

 

[statdad]

Does your first question mean "What is the probability A is the fastest runner among A, B, C, and D?" (with runner E not coming into play at all?) If so,
I know you typed their distributions, but what does the information tell you about the distributions for A, B, C, and D?

 

[thereddevils]

Does your first question mean "What is the probability A is the fastest runner among A, B, C, and D?" (with runner E not coming into play at all?) If so,
I know you typed their distributions, but what does the information tell you about the distributions for A, B, C, and D?

I thought of this:

P(A-B>0) x P(A-C>0) x P(A-D>0)

and P(A-B>0)=P(A-C>0)=P(A-D>0)

but the problem is the calculation of the adjusted mean and SD. Take A-B for example,

E(A-B)=E(A)-E(B)=0

so P(A-B>0)=P(Z>0)=0.5 ?

and the required probability would be (0.5)^3 ?

 

[statdad]

Not quite: think this way: A, B, C, D are identically distributed: if you have four identical items, what is the probability of selecting A (as the smallest)?

 

[thereddevils]

Not quite: think this way: A, B, C, D are identically distributed: if you have four identical items, what is the probability of selecting A (as the smallest)?

oops maybe i have been thinking too much. The answer is simply 1/4. Am i right?

 

[statdad]

:). Good